How find this nice limit It is well kown this following
$$\lim_{x\to+\infty}\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x=\sqrt{ab}(a,b>0)$$
and also kown this general
$$\lim_{x\to+\infty}\left(\dfrac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+a^{\frac{1}{x}}_{n}}{n}\right)^x=\sqrt[n]{a_{1}a_{2}\cdots a_{n}},(a_{i}>0,i=1,2,\cdots,n)$$
and some hours ago,I found this nice and Hard limit
$$\lim_{x\to+\infty}x\left[\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x-\sqrt{ab}\right]=\dfrac{\sqrt{ab}}{8}\left(\ln{a}-\ln{b}\right)^2$$
my proof
$$a^{\frac{1}{x}}+b^{\frac{1}{x}}=e^{\frac{1}{x}\ln{a}}+e^{\frac{1}{x}\ln{b}}=1+\dfrac{1}{x}\ln{a}+\dfrac{1}{2x^2}\ln^2{a}+1+\dfrac{1}{x}\ln{b}+\dfrac{1}{2x^2}\ln^2{b}+o(1/x^2)$$
then
\begin{align*}
\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x&=\left[1+\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)+o(\frac{1}{x^2})\right]^x\\
&\approx e^{x\ln{\left(1+\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)\right)}}\\
&\approx e^{x\left(\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)\right)}\\
&=\cdots\cdots\\
&\approx \sqrt{ab}\left(1+\dfrac{1}{8x}\left(2\ln^2{a}+2\ln^2{b}-\ln^2{(ab)}\right)\right)
\end{align*}
so
$$\lim_{x\to+\infty}x\left[\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x-\sqrt{ab}\right]=\dfrac{\sqrt{ab}}{8}\left(\ln{a}-\ln{b}\right)^2$$
My question:
$$\lim_{x\to+\infty}x\left[\left(\dfrac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+a^{\frac{1}{x}}_{n}}{n}\right)^x-\sqrt{a_{1}a_{2}\cdots a_{n}}\right]=?$$
 A: We can compute
$$\lim_{x\to +\infty} x\left[\left(\frac{a_1^{1/x} + a_2^{1/x} + \dotsb + a_n^{1/x}}{n}\right)^x - \sqrt[n]{a_1a_2\dotsb a_n}\right]$$
in the same way. For brevity, write $\lambda_\nu = \log a_\nu$, then we have
$$\begin{align}
\frac1n\sum_{\nu=1}^n a_\nu^{1/x} &= \frac1n\sum_{\nu=1}^n \exp\left(\frac1x\lambda_\nu\right)\\
&= \frac1n\sum_{\nu=1}^n \left(1 + \frac{\lambda_\nu}{x} + \frac{\lambda_\nu^2}{2x^2} + O(x^{-3})\right)\\
&= 1 + \frac{1}{nx}\sum_{\nu=1}^n\lambda_\nu + \frac{1}{2nx^2}\sum_{\nu=1}^n\lambda_\nu^2 + O(x^{-3}).
\end{align}$$
Write $\mu := \frac1n\sum\limits_{\nu=1}^n\lambda_\nu$ and $s = \frac1n\sum\limits_{\nu=1}^n \lambda_\nu^2$, so we have
$$A := \frac1n\sum_{\nu=1}^n a_\nu^{1/x} = 1 + \frac{\mu}{x} + \frac{s}{2x^2} + O(x^{-3}).$$
Thus
$$\begin{align}
\log A &= \frac{\mu}{x} + \frac{1}{2x^2}\left(s - \mu^2\right) + O(x^{-3}),\\
x\log A &= \mu + \frac{1}{2x} (s-\mu^2) + O(x^{-2}),\\
A^x - \sqrt[n]{a_1a_2\dotsb a_n} &= \exp \left(x\log A\right) - e^\mu\\
&= e^\mu\left(\exp (x\log A - \mu) - 1\right)\\
&= e^\mu\left(\exp \left(\frac{s-\mu^2}{2x} + O(x^{-2}) \right)-1\right)\\
&= e^\mu\left(\frac{s-\mu^2}{2x} + O(x^{-2})\right)\\
x\left(A^x - e^\mu\right) &= e^\mu\left( \frac{s-\mu^2}{2} + O(x^{-1})\right),
\end{align}$$
whence
$$\begin{align}
\lim_{x\to +\infty}&\; x\left[\left(\frac{a_1^{1/x} + a_2^{1/x} + \dotsb + a_n^{1/x}}{n}\right)^x - \sqrt[n]{a_1a_2\dotsb a_n}\right]\\
&= e^\mu\frac{s-\mu^2}{2}\\
&= \sqrt[n]{a_1a_2\dotsb a_n}\frac{n\sum\limits_{\nu=1}^n \log^2 a_\nu - \left(\sum\limits_{\nu=1}^n \log a_\nu\right)^2}{2n^2}.
\end{align}$$
